Question

Find the values of the other five trigonometric functions in each of the following:
(i) $\cot \mathrm{\theta }=\frac{12}{5},$ θ in quadrant III
(ii) $\cos \mathrm{\theta }=-\frac{1}{2},$ θ in quadrant II
(iii) $\tan \mathrm{\theta }=\frac{12}{5},$ θ in quadrant III
(iv) $\sin \mathrm{\theta }=\frac{3}{5},$ θ in quadrant I

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}: \\ \cot \theta =\frac{12}{5}\mathrm{and}\theta \text{are}\mathrm{in}\text{the third quadrant.} \\ \mathrm{In}\text{the third quadrant,}\tan \theta \mathrm{and}\cot \theta \mathrm{are}\mathrm{positive} \\ \text{A}\mathrm{nd},\sin \theta ,\cos \theta ,\sec \theta \mathrm{and}\mathrm{cosec}\theta \mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \tan \theta =\frac{1}{cot\theta }=\frac{1}{{\displaystyle \frac{12}{5}}}=\frac{5}{12} \\ \cos ec\theta =-\sqrt{1+co{t}^2\theta }=-\sqrt{1+{\left({\displaystyle \frac{12}{5}}\right)}^2}=-\frac{13}{5} \\ \sin \theta =\frac{1}{\cos ec\theta }=\frac{1}{-\frac{13}{5}}=-\frac{5}{13} \\ \cos \theta =-\sqrt{1-{\sin }^2\theta }=-\sqrt{1-{\left({\displaystyle \frac{-5}{13}}\right)}^2}=\frac{-12}{13} \\ \mathrm{And},sec\theta =\frac{1}{\cos \theta }=\frac{1}{{\displaystyle \frac{-12}{13}}}=\frac{-13}{12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}: \\ \cos \theta =-\frac{1}{2}\mathrm{and}\theta \text{are in the second quadrant.} \\ \mathrm{In}\text{the second quadrant,}\sin \theta ,\mathrm{and}\mathrm{cosec}\theta \mathrm{are}\mathrm{positive}. \\ \mathrm{And},\tan \theta ,\cot \theta ,\cos \theta \mathrm{and}\sec \theta \mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\left({\displaystyle \frac{-1}{2}}\right)}^2}=\frac{\sqrt{3}}{2} \\ \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\sqrt{3} \end{gathered}$$
$$\begin{gathered} cot\theta =\frac{1}{\tan \theta }=\frac{1}{-\sqrt{3}}=\frac{-1}{\sqrt{3}} \\ sec\theta =\frac{1}{\cos \theta }=\frac{1}{{\displaystyle \frac{-1}{2}}}=-2 \\ \cos ec\theta =\frac{1}{\sin \theta }=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}: \\ \tan \theta =\frac{3}{4}\mathrm{and}\theta \text{are in the third quadrant.} \\ \mathrm{In}\text{the third quadrant,}\tan \theta ,\mathrm{and}\cot \theta \mathrm{are}\mathrm{positive}. \\ \mathrm{And},\sin \theta ,\cos \theta ,\sec \theta \mathrm{and}\mathrm{cosec}\theta \mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore cot\theta =\frac{1}{\tan \theta }=\frac{1}{{\displaystyle \frac{3}{4}}}=\frac{4}{3} \\ sec\theta =-\sqrt{1+{\tan }^2\theta }=-\sqrt{1+{\left({\displaystyle \frac{3}{4}}\right)}^2}=-\frac{5}{4} \\ \cos \theta =\frac{1}{sec\theta }=\frac{1}{-\frac{5}{4}}=-\frac{4}{5} \\ \sin \theta =-\sqrt{1-{\cos }^2\theta }=-\sqrt{1-{\left({\displaystyle \frac{-4}{5}}\right)}^2}=\frac{-3}{5} \\ \cos ec\theta =\frac{1}{\sin \theta }=\frac{1}{-\frac{3}{5}}=-\frac{5}{3} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}: \\ \sin \theta =\frac{3}{5}\mathrm{and}\theta \text{are in the first quadrant.} \\ \mathrm{In}\text{the first quadrant,}\mathrm{all}\mathrm{six}\mathrm{T}-\mathrm{ratios}\mathrm{are}\mathrm{positive}. \end{gathered}$$
$$\begin{gathered} \therefore \cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-{\left({\displaystyle \frac{{\displaystyle 3}}{5}}\right)}^2}=\frac{4}{5} \\ \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{3}{4} \\ cot\theta =\frac{1}{\tan \theta }=\frac{1}{\frac{3}{4}}=\frac{4}{3} \\ sec\theta =\frac{1}{\cos \theta }=\frac{1}{{\displaystyle \frac{4}{5}}}=\frac{5}{4} \\ \cos ec\theta =\frac{1}{\sin \theta }=\frac{1}{{\displaystyle \frac{3}{5}}}=\frac{5}{3} \end{gathered}$$